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Given a smooth map $F:\mathbb{R}^n\to\mathbb{R}^n$ with $\det(DF)=0$ is it possible to approximate it by smooth maps $F_i:\mathbb{R}^n\to\mathbb{R}^n$ with $\det(DF_i)\neq 0$ (maybe uniformly)?

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  • $\begingroup$ Is there any requirement for the approximation? For example, $n=2$. $F = (f(x,y), g(x,y))$, $\det (\mathrm{D}F) = 0$ means $f_x \cdot g_y = f_y \cdot g_x$. $\endgroup$
    – River Li
    Apr 20, 2020 at 3:52
  • $\begingroup$ I don't have any requirements for the approximation. This question was motivated by a book I'm reading where they say a proof can be reduced to the invertible case, without saying anything else. $\endgroup$
    – Sak
    Apr 20, 2020 at 16:09
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    $\begingroup$ By $\det DF=0$ do you mean $0$ everywhere? $\endgroup$
    – zhw.
    Apr 22, 2020 at 16:37
  • $\begingroup$ @zhw. If $\det(\mathrm{D}F)= 0$ means zero everywhere, for example, $n=2$, $F=(f(x,y), g(x,y))$ with $\det(\mathrm{D}F)= 0$, i.e., $\frac{\partial f}{\partial x} \cdot \frac{\partial g}{\partial y} = \frac{\partial f}{\partial y} \cdot \frac{\partial g}{\partial x}$ for all $(x,y)\in \mathbb{R}^2$, I guess, the question is: For any given $\epsilon > 0$, can we always find $F_1 = (f_1(x,y), g_1(x,y))$ such that $\det(\mathrm{D}F_1)\ne 0$ for any $(x,y)\in \mathbb{R}^2$, satisfying $|f_1(x,y) - f(x,y)| < \epsilon$ and $|g_1(x,y) - g(x,y)| < \epsilon$ for all $(x,y)\in \mathbb{R}^2$? $\endgroup$
    – River Li
    Apr 26, 2020 at 4:13
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    $\begingroup$ @RiverLi Yes, that is the interpretation I was asking about. $\endgroup$
    – zhw.
    Apr 26, 2020 at 16:37

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If $DF$ is diagonalizable, an easy way could be to add $\varepsilon(x_1,\dots,x_n)$ and let $\varepsilon\to 0$. Otherwise, you should use the fact that nondiagonalizable matrices can be approximated by diagonalizable ones and adjust your $F$ appropriately to match the nondiagonalizable matrices. They have a nice explicit form: you can prove this by showing every matrix is similar to an upper triangular matrix.

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    $\begingroup$ But $\epsilon$ depends on $x$, right? I agree that your method provides a map such that at one $x \in \mathbb{R}^n$ the jacobian is invertible. But can we do that for all $x \in \mathbb{R}^n$? Maybe I didn't understand really well the question.. $\endgroup$
    – jvc
    Apr 25, 2020 at 21:57
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    $\begingroup$ Yes, you're right, I was assuming DF is zero at a point. If DF=0 everywhere, I think you could approximate by Schwartz functions in each component for uniformity, choosing them so their derivative dorsn't all vanish in the same place. $\endgroup$ Apr 26, 2020 at 22:26

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